Does there exist a conformal $\phi: D\rightarrow\Omega\cup\{\infty\}$?

Question

Let $\gamma$ be a Jordan curve and $\Omega$ the unbounded connected component of $\mathbb{C}\setminus\gamma$. $\Omega$ is not simply connected in $\mathbb{C}$, but $\Omega\cup\{\infty\}$ is simply connected in $\mathbb{C}\cup\{\infty\}$. Does there exist a conformal map $\phi: D\rightarrow\Omega\cup\{\infty\}$, where $D$ is the open unit disk? I assume that if such a $\phi$ exists we could also require $\phi(0)=\infty$. Also, can $\phi$ be extended as homeomorphism in $\overline{D}$?

Answer 1

The existence of a conformal map $\phi : D \to \Omega \cup \{\infty\}$ follows from the Riemann mapping theorem. $\Omega \cup \{\infty\}$ is a simply connected proper subdomain of $\mathbb{C}$: just use inversion of the Riemann sphere in a point of $\Omega$. Being conformally equivalent to a proper subdomain of $\mathbb{C}$, $\Omega \cup \{\infty\}$ is not conformally equivalent to $\mathbb{C}$ itself. Thus it is conformally equivalent to $D$, by the Riemann mapping theorem.